115年沙崙高中教甄-數學詳解

臺南市立沙崙國際高級中學高中部115學年度第1次教師甄選

一、 填充題(60%)

解答:

$$兩圖形\cases{y=|\log x|\\ y=ax+b} 有三個交點的情形\\ 假設三交點的x坐標分別為k,2k,4k(k\gt 0) \Rightarrow k\lt 1\lt 2k\lt 4k \Rightarrow \cases{-\log k=ak+b\\ \log(2k)=2ak+b\\ \log(4k)=4ak+b} \\ \Rightarrow k^2={\sqrt 2\over 2} =2^{-1/2}\Rightarrow k=2^{-1/4} \Rightarrow a={1\over 2k}\log 2 \Rightarrow -\log 2^{-1/4}={1\over 2}\log 2+b \Rightarrow b=-{1\over 4}\log 2 \\ \Rightarrow b=\log 2^{-1/4} \Rightarrow 10^b= 2^{-1/4}=2^{3/4}/2 = \bbox[red, 2pt]{\sqrt[4]8\over 2}$$

解答:$$從 9 個空位中，隨便挑 $3$ 個位子塗紅色，有C^9_3種方法;再從剩下的6個空位挑3個塗黃色\\有C^6_3種方法;最後剩下3個空格塗綠色有C^3_3種方法;總共有C^9_3C^6_3C^3_3=1680種方法\\ 賓果有8條，從8條中選一條(8種選法)塗紅色，剩下6格塗3格黃色,3格綠色，有{6!\over 3!3!}=20種\\ 因此紅色達成賓果有8\times 20=160種,記為n(R)=160; 同理,n(Y)=n(G)=160\\ 考慮同時有兩條直線的情形:\cases{兩條平行的橫列:3條中選2條，有3種組合 \\兩條平行的直行:3條中選2條，有3種組合 } \\ \qquad \Rightarrow 共有3+3=6對不相交的直線; 6對直線挑一對，一條塗紅、一條塗黃色, 剩下3空格塗綠 \\ \qquad \Rightarrow 兩種顏色同時連線:n(R\cap Y)=6\times2(紅黃或黃紅)=12,同理n(R\cap G) =n(Y\cap G)=12 \\ 同時有三條直線的情形:\cases{三條橫線:3!=6\\ 三條直行:3!=6} \Rightarrow n(R\cap Y\cap G)=6+6=12 \\ 因此n(R\cup Y\cup G) =n(R)+n(Y)+n(G)-n(R\cap Y) -R(Y\cap G)-R(Y\cap R)+ n(R\cap Y\cap G) \\=480-36+12=456 \Rightarrow 機率={456\over 1680} =\bbox[red, 2pt]{19\over 70}$$

解答:$$\cases{f(x,y) =x^3+y^3-3xy \\g(x,y)=x^2+y^2-4=0} \Rightarrow \cases{f_x=\lambda g_x\\ f_y=\lambda g_y} \Rightarrow \cases{3x^2-3y= \lambda(2x) \\3y^2-3x=\lambda(2y)} \;兩式相除\Rightarrow {x^2-y\over y^2-x}={x\over y} \\ \Rightarrow x^2y-y^2=xy^2-x^2 \Rightarrow x^2-y^2+x^2y-xy^2=0\Rightarrow (x-y)(xy+x+y) =0 \\ \textbf{Case I }x=y \Rightarrow 2x^2=4\Rightarrow x=\pm \sqrt 2 \Rightarrow y=\pm \sqrt 2 \Rightarrow \cases{f(\pm\sqrt 2, \pm\sqrt 2)= 4\sqrt 2-6 \\f(\pm \sqrt 2,\mp\sqrt 2) =6} \\\textbf{Case II }xy=-(x+y) \Rightarrow x^2+y^2-4=(x+y)^2-2xy-4=(x+y)^2+2(x+y)+4=0 \\ \qquad \Rightarrow \cases{x+y=-1+ \sqrt 5 \Rightarrow xy=1-\sqrt 5 \Rightarrow f(x,y)=(x+y)^3-3xy(x+y)-3xy= 5\sqrt 5-1\\x+y=-1-\sqrt 5 \Rightarrow xy=1+ \sqrt 5 \Rightarrow f(x,y)= -1-5\sqrt 5} \\ \Rightarrow 最大值為 \bbox[red, 2pt]{5\sqrt 5-1}$$

解答:$$P(x)={x^{10}-1\over x-1} =x^9+x^8+\cdots+1 =(x-z)(x-z^2)\cdots (x-z^9) \\ \Rightarrow P'(x)=9x^8 +8x^7+ \cdots 2x+1 \Rightarrow \cases{P(1)=10\\ P'(1)=45} \\ \Rightarrow \ln P(x)= \sum_{k=1}^9 \ln(x-z^k) \Rightarrow {P'(x) \over P(x)}= \sum_{k=1}^9 {1\over x-z^k} \\取x=1 \Rightarrow \sum_{k=1}^9 {1\over 1-z^k} ={P'(1) \over P(1)}= {45\over 10} = \bbox[red, 2pt]{9\over 2}$$

解答:$$f(x)=x^n =(x-2)(x-3)(x-4)Q(x)+ax^2+bx+c \Rightarrow \cases{f(2)=2^n= 4a+2b+c\\ f(3)=3^n=9a+3b+c\\ f(4)=4^n =16a+4b+c} \\ \Rightarrow \begin{bmatrix}4& 2& 1 \\ 9& 3& 1 \\16& 4& 1\end{bmatrix} \begin{bmatrix}a\\ b\\c \end{bmatrix} = \begin{bmatrix}2^n\\ 3^n\\4^n \end{bmatrix} \Rightarrow \begin{bmatrix}a\\ b\\c \end{bmatrix} =\begin{bmatrix}4& 2& 1 \\ 9& 3& 1 \\16& 4& 1\end{bmatrix} ^{-1} \begin{bmatrix}2^n\\ 3^n\\4^n \end{bmatrix}= \begin{bmatrix} {1\over 2} \left(  4^n+2^n-2\cdot 3^n\right)  \\ {1\over 2} \left( -5\cdot 4^n-7\cdot 2^n+12\cdot 3^n \right)\\  \left( 3\cdot 4^n+6\cdot 2^n-8\cdot 3^n \right) \end{bmatrix} \\ \Rightarrow {R(-1) \over R(0)} ={a-b+c\over c} ={6\cdot 4^n+10\cdot 2^n-15\cdot 3^n\over 3\cdot 4^n+6\cdot 2^n-8\cdot 3^n} \Rightarrow \lim_{n\to \infty} {R(-1) \over R(0)} ={6\over 3}=\bbox[red, 2pt]2$$

解答:$$$$

解答:$$\sqrt{a_n}= {\sqrt{a_{n-1}}+ \sqrt{a_{n+1}} \over 2} \Rightarrow 2\sqrt{a_n} =\sqrt{a_{n-1}}+ \sqrt{a_{n+1}} \Rightarrow \sqrt{a_{n+1}}-\sqrt{a_n} =\sqrt{a_n}-\sqrt{a_{n-1}} \\ \Rightarrow \langle \sqrt{a_n} \rangle 為一等差數列, 且\cases{\sqrt{a_1}= \sqrt 2\\ \sqrt{a_2} =\sqrt{8}=2\sqrt 2} \Rightarrow 公差d=\sqrt 2 \Rightarrow \sqrt{a_n}=\sqrt 2+(n-1)\sqrt 2= n\sqrt 2 \\ \Rightarrow a_n=2n^2 \Rightarrow a_{300} =2\cdot 300^2= \bbox[red, 2pt]{180000}$$

解答:$$底面正方形對角線長為6\sqrt 2\Rightarrow 底面中心點到底面四個頂點均為3\sqrt 2 \\ \Rightarrow 四角錐的高h\Rightarrow h^2+(3\sqrt 2)^2=(3\sqrt 6)^2 \Rightarrow h=6 \\ 假設底面在平面z=0,底面中心點為(4,4,0) \Rightarrow A(4,4,6) \Rightarrow 立方體頂面在平面z=8上\\ \Rightarrow 頂面四個頂點為(0,0,8),(8,0,8),(0,8,8),(8,8,8) 且B是其中一點 \\ \Rightarrow A至四個頂點距離均為\sqrt{4^2+4^2+2^2}=6 \Rightarrow \overline{AB}= \bbox[red, 2pt]6$$

解答:$$假設\cases{事件O:特沙拉遇到障礙物\\ 事件A:系統判定左側有障礙物} \Rightarrow \cases{P(O)=0.3 \Rightarrow P(O')=1-0.3=0.7\\ P(A|O) =25\%=0.25 \\ P(A|O')= 0.4\% =0.004} \\ 欲求P(O|A)= {P(O\cap A) \over P(A)} ={P(A|O) \cdot P(O) \over P(A|O)\cdot P(O)+P(A|O') \cdot P(O')} \\={0.25\times 0.3\over 0.25\times 0.3+ 0.004\times 0.7}={0.075\over 0.0778} ={375\over 389} \approx \bbox[red, 2pt]{0.964}$$

解答:$$\cases{a_{n+1} =a_n-2b_n\\ b_{n+1}= a_n+4b_n}\; 兩式相加\Rightarrow a_{n+1}+ b_{n+1}=2(a_n+b_n) \Rightarrow \langle a_n+b_n \rangle 為公比數列\\ a_1+b_1=1+1=2 \Rightarrow a_n+b_n= \bbox[red, 2pt]{2^n}$$

解答:$$\alpha^2+4\beta^2+5 \gamma^2-2\alpha \gamma-8\beta \gamma =(\alpha^2-2\alpha r+\gamma^2) +4(\beta-2\beta \gamma+ \gamma^2) = (\alpha-\gamma)^2+4(\beta-\gamma)^2=0 \\ \Rightarrow |\alpha-\gamma|^2=4|\beta-\gamma|^2 \Rightarrow 6^2 =4|\beta-\gamma|^2  \Rightarrow |\beta-\gamma|=3 \\ 由於 (\alpha-\gamma)^2+4(\beta-\gamma)^2=0 \Rightarrow (\alpha-\gamma)^2=-4(\beta-\gamma)^2  \Rightarrow \alpha-\gamma= \pm2i(\beta-\gamma) \\ \Rightarrow \alpha-\beta=(\alpha-\gamma)-(\beta-\gamma) =\pm 2i(\beta-\gamma)-(\beta-\gamma) =(-1\pm 2i)(\beta-\gamma) \\ \Rightarrow |\alpha-\beta|=|-1\pm 2i||\beta-\gamma| =\sqrt 5\cdot 3= \bbox[red, 2pt]{3\sqrt{5}}$$

解答:$$ \begin{vmatrix} 0& x&y&-z\\ -x& 0& 1& -1\\-y& -1& 0& 1\\z& 1&-1&0 \end{vmatrix} =-x \begin{vmatrix} -x& 1& -1\\-y& 0& 1\\z& -1& 0 \end{vmatrix}+ y \begin{vmatrix} -x& 0& -1\\-y& -1& 1\\z& 1& 0 \end{vmatrix} +z \begin{vmatrix} -x&0& 1\\-y& -1&0\\z& 1& -1 \end{vmatrix} \\=-x(-y+z-x) +y(y-z+x) +z(-x-y+z)=(x+y-z)^2=25 \Rightarrow x+y-z=\pm 5\\ \Rightarrow \cases{E_1: x+y-z=5\\E_2:x+y-z=-5} \Rightarrow d(E_1,E_2)={10\over \sqrt{3}} =\bbox[red, 2pt]{10\sqrt 3\over 3}$$

二、 計算證明題(20%)

 解答:$$\sum_{k=0}^\infty q^k=1+ q+q^2 +\cdots ={1\over 1-q} \Rightarrow {d\over dq}\sum_{k=0}^\infty q^k= {d\over dq} \left( {1\over 1-q} \right) \Rightarrow \sum_{k=1}^\infty kq^{k-1} ={1\over (1-q)^2} \\ \Rightarrow P(X=k)= q^{k-1}p \Rightarrow E(X)= \sum_{k=1}^\infty kP(X=k) = \sum_{k=1}^\infty kq^{k-1}p ={p\over (1-q)^2} ={p\over p^2} ={1\over p} \\ \Rightarrow E(X)={1\over p}, \bbox[red, 2pt]{故得證}$$

解答:$$E:2x+3y-6z=24 \Rightarrow {x\over 12}+{y\over 8}-{z\over 4}=1 \Rightarrow E與三軸交點\cases{A(12,0,0) \\B(0,8,0) \\ C(0,0,-4)} \\ \Rightarrow OABC為直角四面體 \Rightarrow 外接球半徑R={1\over 2} \sqrt{\overline{OA}^2+ \overline{OB}^2 +\overline{OC}^2} =2\sqrt{14} \\ \Rightarrow 四面體體積={1\over 6} \cdot \overline{OA} \cdot \overline{OB}  \cdot \overline{OC}=64 \Rightarrow 三個直角形面積\cases{\triangle OAB=48\\ \triangle OBC=16\\ \triangle OCA=24} \\\Rightarrow \triangle ABC= \sqrt{48^2+16^2+24^2} =56 \text{（德古阿定理 De Gua's theorem）} \\ \Rightarrow 四面體總表面積=48+16+24+56=144 \Rightarrow 64={1\over 3}\times r\times 144 \Rightarrow r={4\over 3} \\\Rightarrow R+r= \bbox[red, 2pt]{{4\over 3}+2\sqrt{14}}$$

三、 教材教法與課程設計 (20%)

$$\bbox[cyan,2pt]{略}$$

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解題僅供參考，其他教甄試題及詳解